Update Math4202_L30.md

content/Math4202/Math4202_L30.md

@@ -26,3 +26,66 @@ There is a group homomorphism $\Phi:\pi_1(X,x_0)\to H$ making the diagram commut
 
 We may change the base point using conjugations.
 
+<details>
+<summary>Side notes about free product of two groups</summary>
+
+Consider arbitrary group $G_1,G_2$, then $G_1\times G_2$ is a group.
+
+Note that the inclusion map $i_1:G_1\to G_1\times G_2$ is a group homomorphism and the inclusion map $i_2:G_2\to G_1\times G_2$ is a group homomorphism. The image of them commutes since $(e,g_2)(g_1,e)=(g_1,g_2)=(g_1,e)(e,g_2)$.
+
+#### The universal property
+
+Then we want to have a group $G$ such that for all group homomorphism $\phi:G_1\to H$ and $G_2\to H$, such that there always exists a map $\Phi: G\to H$ such that:
+
+- $\Phi\circ i_1=\phi_1$
+- $\Phi\circ i_2=\phi_2$
+
+#### How to construct the free group?
+
+We consider
+
+$$
+G_1*G_2=S=\{g_1h_1g_2h_2:g_1,g_2\in G_1,h_1,h_2\in G_2\}/\sim
+$$
+
+And we set $g_ie_{G_2}g_{i+1}\sim g_ig_{i+1}$ for $g_i\in G_1$ and $g_{i+1}\in G_2$.
+
+And $h_je_{G_1}h_{j+1}\sim h_jh_{j+1}$ for $h_j\in G_2$ and $h_{j+1}\in G_1$.
+
+And we define the group operation
+
+$$
+(g_1 h_1\cdots g_k h_k)*(h_1' g_1'\cdots h_l' g_l')=g_1 h_1\cdots g_k h_k g_1' h_2'\cdots h_l' g_l'
+$$
+
+And the inverse is defined
+
+$$
+(g_1 h_1\cdots g_k h_k)^{-1}=h_k^{-1} g_k^{-1}\cdots h_1^{-1} g_1^{-1}
+$$
+
+And $G=S$ is a well-defined group.
+
+The homeomorphism $G\to H$ is defined as
+
+$$
+\Phi((g_1 h_1\cdots g_k h_k))=\phi_1(g_1)\circ \phi_2(h_1)\circ \cdots \circ \phi_1(g_k)\circ \phi_2(h_k)
+$$
+
+Note $\circ$ is the group operation in $H$.
+
+> Group with such universal property is unique, so we don't need to worry for that too much.
+
+</details>
+
+Back to the Seifert-Van Kampen Theorem:
+
+Let $H=\pi_1(U,x_0)* \pi_1(V,x_0)$.
+
+Let $N$ be the **least normal subgroup** in the free product $H$, containing $i_1(g)i_2(g)^{-1}$, $\forall g\in \pi_1(U\cap V,x_0)$.
+
+Note $i_1(g)\in \pi_1(U,x_0)$ and $i_2(g)\in \pi_1(V,x_0)$. You may think of them as $G_1,G_2$ in the free group descriptions.
+
+#### Seifert-Van Kampen Theorem (classical version)
+
+There is an isomorphism between $\pi_1(U,x_0)* \pi_1(V,x_0)/N$ and $\pi_1(U\cup V,x_0)$.